In ASA system, the side cutting and end cutting edge angles of a sharp turning tool are 45° and 10°, respectively. The feed during cylindrical turning is 0.1 mm/rev. The centre line average surface roughness (in μm, round off to one decimal place) of the generated surface is ___________.

The correct answer is 3.75.

To find the centre line average (CLA) surface roughness of the generated surface when using a sharp turning tool, we can follow these step-by-step calculations:

1. Identify the given parameters:
Side cutting edge angle, $C_s=45°$
End cutting edge angle, $C_e=10°$
Feed rate during turning, $f=0.1\text{mm/rev}$

2. Calculate the peak-to-valley roughness height ($h$):
For a single-point sharp turning tool (with zero nose radius), the peak-to-valley height is given by the formula:
$h=\frac{f}{tan\left(C_s\right)+cot\left(C_e\right)}$
Substituting the given values into the equation:
$h=\frac{0.1}{tan(45°)+cot(10°)}$
We know that:
$tan(45°)=1$
$cot(10°)=\frac{1}{tan(10°)}\approx \frac{1}{0.176327}\approx 5.67128$
Therefore, the peak-to-valley height $h$ is:
$h=\frac{0.1}{1+5.67128}=\frac{0.1}{6.67128}\approx 0.01499\text{mm}$
Converting $h$ from millimeters to micrometers ($\text{μm}$):
$h\approx 0.01499\times 1000\text{μm}=14.99\text{μm}$

3. Calculate the centre line average surface roughness ($R_a$):
For a profile with a sharp tool, the centre line average surface roughness is approximately one-fourth of the peak-to-valley height:
$R_a=\frac{h}{4}$
Substituting the calculated value of $h$:
$R_a=\frac{14.99}{4}$
Rounding to two decimal places, we get:
$R_a=3.75\text{μm}$

Thus, the centre line average surface roughness of the generated surface is 3.75 μm.